# How do you solve for r in - \frac{59}{60} = \frac{1}{6} \(- \frac{4}{3} r-5 )?

Dec 12, 2014

Answer: $r = \frac{27}{40}$

Explanation:

$- \frac{59}{60} = \frac{1}{6} \left(- \frac{4}{3} r - 5\right)$

We can now multiply both sides of the equation by $6$:

$\frac{- 59 \cdot 6}{60} = - \frac{4}{3} r - 5$

We notice that $60 = 6 \cdot 10$ so we can cancel out the $6$ on the left hand side:

$\frac{- 59 \cdot 6}{10 \cdot 6} = - \frac{4}{3} r - 5$

$- \frac{59}{10} = - \frac{4}{3} r - 5$

Now we can add $5$ to both sides to further isolate r:

$- \frac{59}{10} + 5 = - \frac{4}{3} r - 5 + 5$

$- \frac{59}{10} + \frac{50}{10} = - \frac{4}{3} r$

$- \frac{9}{10} = - \frac{4}{3} r$

And then multiply both sides by $\frac{3}{4}$ to get the value of r:

$- \frac{9}{10} \cdot - \frac{3}{4} = - \frac{4}{3} r \cdot - \frac{3}{4}$

$- \frac{9}{10} \cdot - \frac{3}{4} = r$

$\frac{27}{40} = r$

$r = \frac{27}{40}$

Checking work:

$- \frac{59}{60} = \frac{1}{6} \left(- \frac{4}{3} r - 5\right)$

$- \frac{59}{60} = \frac{1}{6} \left(- \frac{4}{3} \cdot \frac{27}{40} - 5\right)$

$- \frac{59}{60} = \frac{1}{6} \left(- \frac{108}{120} - 5\right)$

$- \frac{59}{60} = \frac{1}{6} \left(- \frac{27}{30} - 5\right)$

$- \frac{59}{10} = \left(- \frac{27}{30} - 5\right)$

$- \frac{59}{10} + 5 = - \frac{27}{30}$

$\frac{- 59 + 50}{10} = - \frac{9}{10}$

$- \frac{9}{10} = - \frac{9}{10}$